The Open Door Web Site
HOME PAGE BIOLOGY CHEMISTRY PHYSICS ELECTRONICS HISTORY HISTORY of SCI & TECH MATH STUDIES LEARN FRENCH STUDY GUIDE PHOTO GALLERY
ATOMIC and NUCLEAR ELECTRICITY and MAGNETISM MEASUREMENTS MECHANICS OPTICS PRACTICAL WORK QUESTIONS RELATIVITY THERMAL PHYSICS WAVES
MECHANICS
Google
Custom Search
Gravitational Field Strength and Gravitational Potential
Gravitational Field Strength    
G.F.S at a point is defined as the force per unit mass acting on a small mass placed at that point.  
   
Therefore, the units of G.F.S. are Nkg-1.  
   
However, Newton's second law of motion reminds us that force per unit mass (that is, force divided by mass) is equivalent to acceleration.  
This means that G.F.S and acceleration due to gravity are, in fact, different terms for the same quantity.  
   
G.F.S. at a point a distance r away from a point mass M  
Using Newton's law of gravitation and Newton's second law of motion (see here) we have  
 
This shows that g (or G.F.S.) varies with the inverse square of the distance r.  
   
So a graph of g against r is as shown below.  
 
G.F.S. at a point a distance r away from a the centre of a sphere of radius R and mass M  
   
Outside the sphere, the same inverse square law applies.  
So, outside the sphere, the graph is as shown below.  
   
   
It can be shown that, inside the sphere, the G.F.S. is directly proportional to the distance from the centre.  
So the complete graph is  
   
Note that, the maximum value of g (of G.F.G.), is at the surface of the body.  
   
Variation of g (of G.F.S.) on a line joining the centres of two point masses  
Consider the following situation  
   
How does g vary if point p starts very close to mass m1 and moves to being very close to mass m2?  
   
When very close to m1 that mass "dominates" and so the curve is pretty similar to the first graph, above (though inverted, due to our choice of the positive sense being directed away from m1)  
   
At some point in between the two masses we arrive at a distance where the two fields have equal magnitude but opposite sense, so the G.F.S. equals zero here.  
In the case shown here, this point is nearer to m2 which means that the greater of the two masses is... answer below...  
   
We can easily find the position of this point using:  
 
where D is the total distance between the centres of the two masses.  
   
As we approach m2 that mass starts to "dominate" the field and so, again, we have a similar curve to the first graph in this region.  
Note that, as we approach either mass, the field strength approaches infinity (since these are defined to be point masses having no spacial distribution).  
   
In this case m1 > m2  
   
Gravitational Potential, V  
The gravitational potential at a point in a gravitational field is defined as the work done per unit mass bringing a small mass from infinity to that point.  
   
The units of potential are therefore Jkg-1  
   
Why "from infinity" ?  
By saying "at infinity", we mean far enough away to be able to assume that the force of gravity due to the body being considered is effectively zero.  
   
Once we have reached this situation, we can move a body around without doing any work against gravitational forces.  
For this reason we consider that a body "at infinity" has zero potential.  
Re-phrasing this statement: the zero of gravitational potential is at infinity.  
   
To calculate a potential at a point we therefore use the equation work done equals force multiplied by displacement.  
However, in this case the calculation is complicated by the fact that the force varies as the mass is moved.  
We do know how the force varies with distance (Newton's law of gravitation)  
 
where, in this case, m1 = M, the mass of the body "causing" the field and m2 = m, the small mass we are moving.  
This allows us to find the following equation for work done*  
 
This is the work done moving a mass m, so the work done per unit mass is given by  
 
   
However, as stated above, the zero of potential is taken to be at infinity.  
Given the chance, a body always falls to its lowest state of potential (energy) so we must arrange that the equation for potential is such that, as r decreases, the potential also decreases.  
We can do this by including a negative sign in the above equation.  
Then, as r decreases, V decreases (becomes a greater negative quantity).  
   
Therefore the potential V at a point a distance r away from a point mass (or from the centre of a uniformly distributed spherical mass) M is given by:  
 
This equation is useful in the calculation of a planet's escape velocity.  
   
   
*To obtain this equation, we imagine moving the mass one tiny step at a time, each step being so small that we can assume that the force does not vary appreciably.  
We then add up all these "infinitesimal" quantities of work to find the total.  
The process is called integration.  
SITE MAP
WHAT'S NEW
ABOUT
PRIVACY
COPYRIGHT
SPONSORSHIP
DONATIONS
ADVERTISING
 

© The Open Door Team
2016
Any questions or
problems regarding
this site should be
addressed to
the webmaster

David Hoult 2017

Hosted By
Web Hosting by HostCentric

 
SiteLock
 
 
Mechanics Index Page